Explanation for Question 14 From the Math (No Calc) Section on the 2020 October Sat
So question 14 asks how many solutions this equation has. 2 And whenever we're asked how many solutions are, 3 what is the solution they want us to solve for X? 4 Now, the way this question is written, it's a little bit tough to solve 5 for X because we have 1, 2, 3, 4 different X variables. 6 So the best thing that we can do to start is actually go ahead 7 and distribute everything so that if we distribute this X, 8 we get X squared minus four X. Then if we, 9 uh, foil this out, we do first outer in or last. 10 We'll see that we get X squared minus X plus X minus one. 11 These two will cancel because they are the same thing. Get X squared, 12 minus four, X is equal to X squared minus one. 13 So now I'll try to get everything on one side or the X is 14 on one side and the normal numbers on the other. Now I'll start by 15 subtracting X squared from both sides. And this is a nice case where my 16 X squares just go away. Because if I subtract from one side, 17 I subtract everything. That's on the other side, then everything else just drops down. 18 And if I want to get rid of the negative four, 19 that's attached to the X, I just need to divide by negative four. 20 And I get X is equal to one fourth. And we noticed that when 21 I solve this, I only get one answer, 22 right? I don't get one, two or three answers. 23 I do get one. I don't get two or three answers. 24 And so since I only have one number, it's the one fourth. 25 Then I only have one solution. And our correct answer is B.